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The absolute residue degree is the degree of the residue field over its prime field. For a tower above $p$, degrees multiply: $[k_{\mathfrak{q}}:{\mathbb{F}}_p]=[k_{\mathfrak{q}}:k_{\mathfrak{p}}][k_{\mathfrak{p}}:{\mathbb{F}}_p]$. In a Galois extension over $\mathbb{Q}$, the decomposition group has a cyclic quotient group of this order by its inertia group. This can exclude residue degrees even when the whole extension has that degree: $S_3$ has no cyclic quotient group of order six.

For a finite abelian extension of number fields, local reciprocity maps into the local decomposition groups combine to a continuous map on the idele group: outside finitely many places they kill units, and an idele has nonunit components only finitely often. The Artin reciprocity law is equivalent to the compatibility that this product kills every principal idele. Conversely, restrict a global continuous Artin homomorphism to each single-place embedding of $K_v^{\ast }$. These maps reconstruct the global map, by density of finite-component ideles and triviality on a basic open subgroup of units. Merely having unrelated local maps omits the essential principal-idele compatibility.

A nonidentity involution in the affine group over a finite field ${\mathbb{F}}_{\ell }$, for odd prime $\ell$, has linear coefficient $-1$: a translation has order $\ell$, while $u^2=1$ forces $u=\pm 1$. Its unique fixed point is $b/2$, and its remaining orbits have size two. This orbit count computes primes in an intermediate field as double cosets for an order-two decomposition group acting on the radical roots.

// Target: number-theory.bigb

The polynomialsplits into three linear factors over ${\mathbb{Q}}_2$. Over ${\mathbb{Q}}_5$ it is an Eisenstein polynomial, so adjoining a root gives a totally ramified cubic extension. Its splitting field over ${\mathbb{Q}}_5$ has decomposition group $S_3$ and inertia group $A_3\cong C_3$.

Let $f\in R[X]$ have degree $n$. A useful reduction prime must preserve its degree and make its reduction separable. Thus exclude the leading coefficient and discriminant from $P$; after localizing at those elements one may make $f$ monic. The residue base field is $k=\mathrm{Frac}(R/P)$, not necessarily the ring $R/P$ itself. Then the Galois group of the reduced polynomial embeds in the generic group, with the same action on the corresponding roots.

Here is the unramified reduction of a polynomial Galois group argument. A UFD is integrally closed. Let $E$ be the splitting field over $F$, and use the root algebra $B$ obtained by adjoining all roots after inverting the discriminant. It is finite étale over the resulting normal base: successive simple-root adjunctions are unramified, and the distinct generic components separate over a normal ring. Choose a prime $Q$ above $P$. The residue field $k_Q$ is generated over $k$ by the reduced roots, so it is their splitting field. The decomposition groupacts on $k_Q$. Its residue action is onto $\mathrm{Gal}(k_Q/k)$; this is the local decomposition-group description of the unramified fiber. Equivalently, the geometric points of that fiber form a free transitive Galois-group set, and the commuting residue-field action identifies the residue group with a stabilizer subgroup of the generic group.

The kernel is the inertia group. An element in that kernel fixes every reduced root. Since the discriminant is a unit, distinct roots have distinct reductions; consequently it fixes every original root and is the identity. ThereforeChoosing another prime changes this subgroup by conjugacy, which does not change the cycle types available for computing the generic group.

When $k={\mathbb{F}}_q$, its splitting-field group is generated by Frobenius. A reduced irreducible factor of degree $d_i$ contributes one $d_i$-cycle. Hence a squarefree factorization with degrees $d_1,\dots ,d_r$ proves that the generic group contains that cycle type. Irreducible degree-preserving reduction also proves generic irreducibility by Gauss's lemma: a nontrivial monic factorization would reduce to one. Several good primes, a discriminant-square test and permutation-group arguments can then identify the group. A ramified or degree-dropping prime does not justify the same cycle test.

For a finite point $p$, use the valuation $v_p$ with local parameter $u=t-p$; at infinity use $u=1/t$. Its extensions $w$ to $E$ are permuted transitively by $G$. The completed residue field is $F$, because $F$ is algebraically closed, so the local decomposition group equals its inertia group.

Write $e=e(w/v_p)$. Characteristic zero makes all local ramification tame. One can choose a uniformizer $\pi$ in the completion for whichIndeed, initially $u$ is ${\pi }^e$ times a unit; that unit has an $e$th root in $F[[\pi ]]$, by its algebraically closed residue field and the invertibility of $e$. Absorb it into $\pi$. The local automorphisms are then $\pi \mapsto \zeta \pi$, $\zeta \in {\mu }_e(F)$, and form a cyclic group of order $e$. Define $e_p=e$.

To specify an element rather than a subgroup, fix a compatible choice of primitive roots of unity ${\zeta }_e\in F$. Define ${\tau }_w$ by its uniformizer characterand let $C_p$ be its conjugacy class in $G$. Over $\mathbb{C}$, take ${\zeta }_e=e^{2\pi i/e}$, corresponding to positive local continuation around $p$. At an unramified point, $e_p=1$ and $C_p=\{1\}$.

Changing the uniformizer does not affect this character. If ${\pi }^{\prime }=a\pi$ with $a$ a unit, inertia fixes its residue, soThe character is injective: a nonidentity finite-order automorphism with derivative one would have $\tau (\pi )=\pi +b{\pi }^r+\cdots$ for $r\ge 2$, and its $N$th iterate would have coefficient $Nb$ at that first nonzero displacement. Characteristic zero contradicts finite order. Thus the chosen primitive character identifies a unique generator.

For another place $w^{\prime }=\sigma w$, conjugation gives $D_{w^{\prime }}=\sigma D_w{\sigma }^{-1}$, preserves its order, and carries ${\tau }_w$ to ${\tau }_{w^{\prime }}$, since $\sigma$ fixes the constants and hence ${\zeta }_e$. This proves that $e_p$ and the branch-cycle conjugacy class are independent of the place and uniformizer. Without the stated roots-of-unity convention, only the cyclic inertia-subgroup conjugacy class is intrinsic: replacing ${\zeta }_e$ by another primitive root replaces the generator by a coprime power, which need not lie in the same conjugacy class.

The idele group is the restricted productwith respect to ${\mathcal{O}}_v^{\ast }$ at the finite places. Thus $a=(a_v)$ is an idele if all components are nonzero and $a_v$ is a unit at all but finitely many finite places. Its topology has basic open sets which equal the full unit subgroup outside finitely many places; it is not just the subspace topology of the unrestricted product. The group is locally compact, and the diagonal $K^{\ast }$ consists of the principal ideles. The idèle class group is $C_K=J_K/K^{\ast }$.

The finite valuations give the idele-to-ideal homomorphismIts kernel is the product of the finite unit groups and the infinite multiplicative groups; principal ideles map to principal fractional ideals. Thus further quotienting gives the ideal class group. The product of normalized local moduli defines $|a{|}_J$; at complex places the normalized modulus is the square of the ordinary absolute value. The product formula says that principal ideles have modulus one. These constructions package ideals, units and archimedean conditions into one group.

Here is the precise reciprocity equivalence. For a finite abelian extension $L/K$, the global law is a continuous Artin reciprocity mapwhich kills $K^{\ast }$ and sends a uniformizer idele at an unramified place to arithmetic Frobenius. Its full class-field-theoretic norm statement is $C_K/N_{L/K}{C}_L\cong \mathrm{Gal}(L/K)$. In the local formulation, the mapsare continuous, kill local units when unramified, send a local uniformizer to the same Frobenius, are compatible with restriction in towers, and obeyHere $D_v$ is the decomposition group of any prime above $v$, independent of that choice because the extension is abelian. The usual local norm theorem identifies $K_v^{\ast }/N_{L_w/K_v}{L}_w^{\ast }$ with $D_v$. The principal-idele condition is the essential global compatibility; existence of unrelated local norm isomorphisms alone would not establish global reciprocity.

Given the compatible local maps, defineOnly finitely many factors are nontrivial: there are finitely many ramified places, and an idele is a unit almost everywhere. The product is a homomorphism. Continuity follows by choosing local kernel neighbourhoods at those exceptional places and the full unit groups elsewhere. The boxed compatibility kills $K^{\ast }$, so the map descends to $C_K$. At every unramified place it has the required Frobenius value. The local norm maps also show that idele norms are killed, since the local component of an idele norm is ${\prod }_{w\mid v}{N}_{L_w/K_v}(b_w)$. Together with the class-field norm-index statement this gives the global norm-kernel formulation.

Conversely, embed $K_v^{\ast }$ in $J_K$ with every other component one, and set $r_v=\theta \circ {\iota }_v$. Each restriction is continuous and has the required unramified normalization. An open kernel of the finite-valued global map contains a basic neighbourhood whose unrestricted finite-place factors are full unit groups outside a finite set. Hence those local restrictions kill units. For an arbitrary idele, remove the finitely many exceptional and nonunit components; the remainder lies in that kernel neighbourhood. It follows directly that $\theta$ is exactly the product of its local restrictions, not merely a map agreeing on single-component ideles. Its triviality on $K^{\ast }$ is exactly the boxed product relation. With the decomposition and norm assertions of the reciprocity theorem, these restrictions are the local reciprocity maps into $D_v$ with kernel $N_{L_w/K_v}{L}_w^{\ast }$.

One can explicitly connect this idelic formulation with the classical ideal law without assuming an idelic law in advance. A modulus of a number field records finitely many nonnegative exponents at finite places and a selected subset of real places. For a modulus $\mathfrak{m}$, let $U_{\mathfrak{m}}$ have factors $1+{\mathfrak{p}}_v^{m_v}$ at finite places dividing the modulus, full units at other finite places, positive real numbers at specified real places, and full infinite multiplicative groups otherwise. Weak approximation for number fields supplies a global $a\in K^{\ast }$ making $a^{-1}j$ lie in the specified local factors at the finitely many modulus places. Its valuations outside the modulus define an ideal prime to $\mathfrak{m}$. A different such $a$ changes that ideal by a ray-principal ideal, and every ideal prime to $\mathfrak{m}$ is obtained. HenceAn ideal Artin map killing $P_K(\mathfrak{m})$ therefore extends to $J_K$ via this isomorphism; restricting it gives the local maps just constructed. Conversely, continuous local maps satisfying the product condition have an open product kernel containing some $U_{\mathfrak{m}}$ and hence make the ideal Artin map kill $P_K(\mathfrak{m})$. This proves the idelic local-global reciprocity equivalence at the reciprocity-law level as well as the formal conversion of maps. The arithmetic Frobenius normalization and product relation agree with Milne's local-global reciprocity formulation.

Let $L/K$ be a finite Galois extension of number fields, with Galois group $G$, and let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$. Write its prime ideal factorization asThe Galois group acts transitively on the primes above $\mathfrak{p}$. One proof uses the Chinese remainder theorem. If $\mathfrak{Q}$ were outside the orbit of $\mathfrak{P}$, choose an integral element $a$ which is zero modulo $\mathfrak{Q}$ and one modulo every conjugate of $\mathfrak{P}$. Its field norm ${\prod }_{\sigma \in G}\sigma (a)$ belongs to $\mathfrak{Q}\cap {\mathcal{O}}_K=\mathfrak{p}$, hence to $\mathfrak{P}$. But each factor is one modulo $\mathfrak{P}$, a contradiction. Transitivity also makes all $e_j$ equal to a common ramification index $e$ and all $f_j$ equal to a common residue degree $f$.

Fix $\mathfrak{P}$ above $\mathfrak{p}$. Its decomposition group is the stabilizerThe orbit-stabilizer theorem gives $|D_{\mathfrak{P}}|=|G|/g$. Elements of $D_{\mathfrak{P}}$ act on the residue field $\kappa (\mathfrak{P})$ over $\kappa (\mathfrak{p})$; the group kernel is the inertia groupThus inertia measures the automorphisms invisible after reduction, whereas decomposition measures all automorphisms preserving the chosen prime.

The completion $L_{\mathfrak{P}}/K_{\mathfrak{p}}$ is a Galois extension with group $D_{\mathfrak{P}}$: preservation of $\mathfrak{P}$ lets an automorphism extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The degree of a field extension formula is $[L_{\mathfrak{P}}:K_{\mathfrak{p}}]=ef$. Reduction gives the exact sequenceFor surjectivity, lift a primitive residue-field generator by the Hensel lemma to obtain the maximal unramified subextension of a local field extension of $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. Its automorphisms are the residue-field automorphisms. They extend to $L_{\mathfrak{P}}$ because the latter is normal. Since a finite residue field extension has a cyclic Galois group of order $f$, the exact sequence yieldsThe fixed field of local inertia is the maximal unramified subextension of a local field extension, and the extension above it is totally ramified. Primes in the same global orbit have conjugate decomposition groups and inertia groups, so the orders do not depend on the chosen prime.

If $e=1$, inertia is trivial and $D_{\mathfrak{P}}$ is generated by the unique Frobenius automorphism acting on residues by $z\mapsto z^{N\mathfrak{p}}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its decomposition group is trivial. A totally ramified prime has $g=f=1$ and $I_{\mathfrak{P}}=D_{\mathfrak{P}}=G$; an unramified prime has $I_{\mathfrak{P}}=1$. In an abelian extension, conjugation does nothing, so the associated groups and unramified Artin symbol are independent of the prime above $\mathfrak{p}$.

Write $\alpha ={\lambda }^{1/5}$, $M=\mathbb{Q}(\alpha )$ and $E=\mathbb{Q}({\zeta }_5)$. The Eisenstein criterion at $\lambda$ gives $[M:\mathbb{Q}]=5$, whereas the prime cyclotomic field degree gives $[E:\mathbb{Q}]=4$. The splitting field $F=E(\alpha )$ is a Galois extension of $E$, and an automorphism fixing $E$ sends $\alpha$ to ${\zeta }_5^b\alpha$. Thus its Galois group embeds in $C_5$. It is nontrivial, since $M$ cannot be contained in the degree-four field $E$. Therefore $[F:E]=5$ and $[F:\mathbb{Q}]=20$.

Index the five roots by $j\in {\mathbb{F}}_5$, with root ${\zeta }_5^j\alpha$. Every automorphism has the form ${\zeta }_5\mapsto {\zeta }_5^u$, $\alpha \mapsto {\zeta }_5^b\alpha$ and acts on indices as $j\mapsto uj+b$, with $u\in {\mathbb{F}}_5^{\times }$. This is a faithful inclusion of $G=\mathrm{Gal}(F/\mathbb{Q})$ into the affine group over a finite field, and both groups have order $20$, so it is an isomorphism. This is the affine Galois group of a prime-radical splitting field.

Let $D$ be the specified decomposition group. Its nonidentity element is an involution. If its linear coefficient were $1$, it would be a nontrivial translation, which has order $5$. Hence it acts as $j\mapsto -j+b$. There is exactly one fixed root, at $j=b/2$, and the other four roots form two pairs. By involution orbits in an affine group of odd prime degree, $D$ has three orbits on the five roots. Those roots identify with $G/\mathrm{Gal}(F/M)$, so the double-coset formula gives three primes in $M$.

For $E$, the embeddings identify with $G/\mathrm{Gal}(F/E)\simeq {\mathbb{F}}_5^{\times }$. The image of $D$ is $\{1,-1\}$ and acts by multiplication, giving the two orbits $\{1,-1\}$ and $\{2,-2\}$. ThereforeThe orbit argument counts primes whether or not $p$ is ramified; cyclicity of the given order-two decomposition group does not require that it be generated by an unramified Frobenius element.

Write $G=\mathrm{Gal}(F/K)$ and fix a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$. Suppose there is another prime $\mathfrak{r}$ above $\mathfrak{p}$ outside the $G$-orbit of $\mathfrak{q}$. Distinct nonzero prime ideals in the ring of integers are distinct maximal ideals, hence pairwise comaximal. The Chinese remainder theorem supplies $a\in {\mathcal{O}}_F$ satisfying $a\equiv 0(\mathrm{mod}\mathfrak{r})$ and $a\equiv 1(\mathrm{mod}\sigma \mathfrak{q})$ for every $\sigma \in G$. Its field norm isThe factor $a$ puts $b$ in $\mathfrak{r}$, hence in $\mathfrak{r}\cap {\mathcal{O}}_K=\mathfrak{p}$. But every $\sigma (a)$ is congruent to $1$ modulo $\mathfrak{q}$, since $a\equiv 1(\mathrm{mod}{\sigma }^{-1}\mathfrak{q})$. Thus $b\equiv 1(\mathrm{mod}\mathfrak{q})$, contradicting $b\in \mathfrak{p}\subset \mathfrak{q}$. This proves transitivity of the Galois action on primes without any unramified hypothesis.

Let $D={\mathrm{Stab}}_G(\mathfrak{q})$ be the decomposition group and $H=\mathrm{Gal}(F/L)$. Transitivity identifies the primes of $F$ above $\mathfrak{p}$ with the coset space $G/D$. Two such primes contract to the same prime in $L$ exactly when they are in the same $H$-orbit: apply the transitivity just proved to the Galois extension $F/L$. Therefore primes in an intermediate field as double cosets givesThe second equality follows by inversion of representatives. It allows one to count $D$-orbits on the embeddings of $L$ into $F$, represented by $G/H$. If $H$ is normal, this simplifies to $[G:HD]$; the double-coset formulation also handles nonnormal intermediate fields.

Write $G=\mathrm{Gal}(F/K)$ andEvery member of this decomposition group preserves ${\mathcal{O}}_F$ and $\mathfrak{q}$, so reduction defines a homomorphismIts kernel is the inertia group. We prove surjectivity of decomposition group reduction without presupposing a lift of Frobenius.

If the two residue fields coincide, surjectivity is immediate. Otherwise choose a nonzero $\overline{a}\in k_{\mathfrak{q}}$ generating it over $k_{\mathfrak{p}}$; a generator exists by the primitive element theorem for finite fields. Use the Chinese remainder theorem in ${\mathcal{O}}_F$ to choose $a$ whose residue is $\overline{a}$ at $\mathfrak{q}$ and zero at every other prime above $\mathfrak{p}$. ConsiderIts coefficients are integral and fixed by $G$, hence lie in ${\mathcal{O}}_K$: an element of $K$ integral over ${\mathcal{O}}_K$ belongs to this integrally closed ring of integers.

Reduce at $\mathfrak{q}$. When $\sigma \notin D_{\mathfrak{q}}$, the chosen residue of $a$ at ${\sigma }^{-1}\mathfrak{q}$ is zero, so $\sigma a$ reduces to zero at $\mathfrak{q}$. When $\sigma \in D_{\mathfrak{q}}$, its residue is $\overline{\sigma }(\overline{a})$, which is nonzero. ThereforeAlthough this product is written in $k_{\mathfrak{q}}[T]$, its coefficients belong to $k_{\mathfrak{p}}$. Put $q=|k_{\mathfrak{p}}|$. The finite-field Frobenius automorphism $b\mapsto b^q$ therefore permutes its nonzero roots. In particular, ${\overline{a}}^q$ is $\overline{\sigma }(\overline{a})$ for some $\sigma \in D_{\mathfrak{q}}$. Since $\overline{a}$ generates $k_{\mathfrak{q}}$, that $\overline{\sigma }$ agrees with Frobenius on the whole field. The Galois group of a finite field extension is cyclic and generated by Frobenius, so its generator lies in the image. Thus

Write $\Gamma =\mathrm{Gal}(K_{\infty }/K)\cong {\mathbb{Z}}_p$. Its only finite subgroup is trivial. In particular, a real place cannot acquire complex inertia in this Zp-extension: the possible nontrivial inertia at a real place has order two. Thus every infinite place splits in the tower.

Suppose that no finite prime ramified. Every finite layer would then be an abelian everywhere unramified extension of number fields, including splitting at real places. Every layer would lie in the ordinary Hilbert class field of $K$, a finite extension. Their degrees are unbounded, a contradiction. Therefore some finite prime must ramify.

Let $v$ have residue characteristic $\ell \ne p$. Apply local class field theory to the corresponding decomposition group, a closed subgroup of $\Gamma$. The image of the unit group ${\mathcal{O}}_{K_v}^{\times }$ is the inertia group. Its first principal-unit subgroup is a pro-l group, so its continuous image in the pro-p group $\Gamma$ is trivial. The residue-unit quotient is the finite group ${\mathbb{F}}_{q_v}^{\times }$. Its image is finite, and is therefore also trivial in the torsion-free group ${\mathbb{Z}}_p$. The entire inertia group is trivial. HenceEquivalently, the tame ramification relation with a Frobenius element would force a tame inertia generator to satisfy ${\tau }^{q_v-1}=1$, which is impossible nontrivially in ${\mathbb{Z}}_p$.

For the final assertion, the cyclotomic Zp-extension of $K$ is $K\mathbb{Q}[p^{\infty }]$. Fix $v\mid p$ and put $E=K_v$, a finite extension of ${\mathbb{Q}}_p$. Let $C_n$ be the completion of $\mathbb{Q}[p^n]$ at its unique prime over $p$. This is a totally ramified extension of ${\mathbb{Q}}_p$ of degree $p^n$. In a compatible local algebraic closure,Here the inequality follows from the tower through $C_n$. The right side is unbounded. These composita occur among the completions of the cyclotomic tower over $K$, so $v$ has unbounded ramification index, in particular nontrivial inertia group. ThusThis argument allows an arbitrary finite intersection between $K$ and the rational cyclotomic tower; it does not assume disjointness.

For a direct check of unramified behavior, the cyclotomic polynomial ${\Phi }_{25}$ divides $X^{25}-1$. Modulo $7$, the latter has derivative $25X^{24}=4X^{24}$ and has no common root with that derivative. Thus ${\overline{\Phi }}_{25}$ is separable, so $7$ does not divide its polynomial discriminant. The field discriminant divides this polynomial discriminant, and therefore $7$ is unramified in $\mathbb{Q}({\zeta }_{25})$. The Frobenius automorphism satisfiesIndeed its residue action is seventh powering. The reduced $25$th roots of unity are distinct because $7\nmid 25$, so the residue congruence identifies the exponent uniquely modulo $25$.

The Galois group is $(\mathbb{Z}/25\mathbb{Z}{)}^{\times }$, of order $\phi (25)=20$. Since $7^2=49\equiv -1(\mathrm{mod}25)$, $7^4\equiv 1$ but $7^2\not\equiv 1$. Consequently the Frobenius automorphism has order four. An unramified decomposition group is generated by that element, and the orbit-stabilizer theorem givesThere are therefore five primes above $7$, each with residue field ${\mathbb{F}}_{7^4}$.

Let $F/K$ be a finite Galois extension, choose a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$, and write $k_{\mathfrak{q}}$ and $k_{\mathfrak{p}}$ for the residue fields. The decomposition group and inertia group areThe Frobenius automorphism on the finite field $k_{\mathfrak{q}}$ is $x\mapsto x^{N\mathfrak{p}}$. Its inverse image defines a distinguished coset in $D_{\mathfrak{q}}/I_{\mathfrak{q}}$; a Frobenius element is a lift of that coset. At an unramified prime, $I_{\mathfrak{q}}=1$ and the lift is unique. Changing the prime above $\mathfrak{p}$ conjugates it, so a Frobenius conjugacy class is intrinsically attached to $\mathfrak{p}$. These are arithmetic Frobenius conventions.

The decomposition group of the chosen prime is its stabilizer under the Galois group:The transitivity of the Galois action on primes identifies the primes of $M$ above $\mathfrak{p}$ with $G/D$ through $gD\mapsto g\mathfrak{P}$.

Put $H=\mathrm{Gal}(M/L)$. Contraction of prime ideals to ${\mathcal{O}}_L$ is constant on every $H$-orbit, because every element of $H$ fixes $L$ pointwise. It is surjective onto the primes of $L$ above $\mathfrak{p}$, since prime ideals extend in the integral extension ${\mathcal{O}}_M/{\mathcal{O}}_L$. Moreover, two primes of $M$ contracting to the same prime of $L$ are conjugate by $H$: the extension $M/L$ is Galois and the same transitivity theorem applies to it.

Thus the fibers of contraction are exactly the $H$-orbits on $G/D$. These orbits are the double cosets $H\backslash G/D$, proving the primes in an intermediate field as double cosets correspondenceExplicitly, replacing $g$ by $hgd$ does not change the contracted prime. Conversely, if $g\mathfrak{P}$ and $g^{\prime }\mathfrak{P}$ contract to the same prime, some $h\in H$ sends the former to the latter, so $g^{\prime -1}hg\in D$ and $H{g}^{\prime }D=HgD$. This also verifies the stated order of the two sides of the double coset.

Write $G=\mathrm{Gal}(L/K)$. The decomposition group and inertia group at $q$ areThe transitivity of the Galois action on primes gives $G/D_{q/p}$ as the set of primes above $p$. Completion induces the canonical isomorphismand reduction gives the exact sequenceConsequently $|I_{q/p}|=e_{q/p}$, $|D_{q/p}/I_{q/p}|=f_{q/p}$ and, if $g$ is the number of primes above $p$,

Let $G=\mathrm{Gal}(L/K)$ and let $\mathcal{P},\mathcal{Q}$ lie over $\mathfrak{p}$. These primes define two extensions to $L$ of the $\mathfrak{p}$-adic absolute value on $K$. The conjugacy of extensions of a valuation to a normal extension says that some $\sigma \in G$ carries the first extension to the second. Equivalently, $\sigma \mathcal{P}=\mathcal{Q}$. This proves the transitivity of the Galois action on primes.

The decomposition group is the stabilizerEach such automorphism extends continuously to the completions, and restriction gives the decomposition group of a prime and local Galois group isomorphism

For the splitting field $L=\mathbb{Q}(\sqrt[3]{7},{\zeta }_3)$ of $X^3-7$, the global Galois group is $S_3$. Since $7\equiv 1(\mathrm{mod}3)$, the field ${\mathbb{Q}}_7$ contains ${\zeta }_3$. The local splitting field is therefore ${\mathbb{Q}}_7(\sqrt[3]{7})$, an Eisenstein, totally ramified cyclic extension of degree three. Hence there areprimes of $L$ above $7$, and the decomposition group of each is the normal subgroup